Let $h(x)=\begin{cases} \dfrac{\sqrt{x+26}-5}{x+1}&\text{for }x\geq -26, x\neq -1 \\\\ k&\text{for }x=-1 \end{cases}$ $h$ is continuous for all $x>-26$. What is the value of $k$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $0$ (Choice C) C $\dfrac{1}{10}$ (Choice D) D $-\dfrac{1}{5}$
Answer: $\dfrac{\sqrt{x+26}-5}{x+1}$ is continuous for all $x>-26$ other than $x=-1$, which means $h$ is continuous for all $x>-26$ other than $x=-1$. In order for $h$ to also be continuous at $x=-1$, the following equality must hold: $\lim_{x\to -1}h(x)=h(-1)$ Since $h(-1)=k$, we will obtain the above equality by letting $k=\lim_{x\to -1}h(x)$. So let's find $\lim_{x\to -1}h(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to -1}h(x) \\\\ &=\lim_{x\to -1}\dfrac{\sqrt{x+26}-5}{x+1} \gray{\text{This is the rule for }x\neq -1} \\\\ &=\lim_{x\to -1}\dfrac{\sqrt{x+26}-5}{x+1}\cdot\dfrac{\sqrt{x+26}+5}{\sqrt{x+26}+5} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to -1}\dfrac{x+26-5^2}{(x+1)(\sqrt{x+26}+5)} \gray{\text{Simplify}} \\\\ &=\lim_{x\to -1}\dfrac{\cancel{x+1}}{\cancel{(x+1)}(\sqrt{x+26}+5)} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to -1}\dfrac{1}{(\sqrt{x+26}+5)} \\\\ &\text{(This is allowed because }x\neq -1) \\\\ &=\dfrac{1}{\sqrt{-1+26}+5} \gray{\text{Direct substitution}} \\\\ &=\dfrac{1}{10} \end{aligned}$ We obtained that if we set $k=\dfrac{1}{10}$, then $\lim_{x\to -1}h(x)=h(-1)$, which makes $h$ continuous at $x=-1$. Since we already saw that $h$ is continuous for any other $x>-26$, we can determine that it's continuous for all $x>-26$. In conclusion, $k=\dfrac{1}{10}$.